Given the vector M = (asinx, cosx), n = (SiNx, bcosx), where a, B, X ∈ R, Let f (x) = m * n satisfy f (π / 6) = 2, and the image of F (x) is symmetric with respect to the line x = π / 3 ① (2) if the equation f (x) + log2k = 0 about X always has a real number solution in the interval [0, π / 2], find the value range of the real number K

① By using the product formula, we get: F (x) = asin ^ x + bcos ^ x into f (π / 6) = 2: asin ^ (π / 6) + bcos ^ (π / 6) = A / 4 + 3B / 4 = 2A + 3B = 8 (1) by using the image symmetry about the straight line x = π / 3: F (π / 3 - π / 3) = f (π / 3 + π / 3) f (0) = f (2 π / 3) B = 3A / 4 + B / 4 a = B (2) (1) (2) simultaneous solution
Sister Xi.! You are caught by me.!! Follow up: how annoying! It means that you have come to investigate this topic! "The image of F (x) is symmetric with respect to the straight line x = π / 3". I forgot how to use this condition to ask. Ouch, I didn't do Tat well in my homework
Let m = (asinx, cosx), n = (SiNx, bsinx), where a, B, X ∈ R. if f (x) = m · n satisfies f (π 6) = 2, and the image of F (x) is symmetric with respect to the straight line x = π 3. (I) find the value of a, B; (II) if the equation f (x) + log2k = 0 with respect to x always has a real solution in the interval [0, π 2], find the value range of real number K
(I) f (x) f (x) f (x) = n (1) f (x) = n (n = n = n = n = n = n = n (1-cos2x) + b2sin2x by F (π 6) = 2, a + 3B = 8 (a + 3B = 8) the image of (1) (1 \\58 (1) for the image of (1) f (x) (a + 3B = 8) the image of (1 b = 8) (1) for the image of the image of (1) (a + 3B = 8) for (a) by (I) by (I) from (I) for (I) we (f (x) f (x) = 1-cos2x (1-cos2x + 3s2x2x2x2x2x + 3 (2x2x2x2x2x = 2) = 2 = 2 (2x = 2 (2x (2x (2-π (2x (2x (2x) - π (2x) if ∵ f (x) + log2k = 0, then ∵ f (x) + log2k = 0 The results show that f (x) = - log2k has a solution, and the solution is 18 ≤ K ≤ 1, that is, K ∈ [18,1]
The known function f (x) = (cosx) ^ 2 + asinx-2a-2
(1) When a = - 2, find the value of X satisfying f (x) = 0
(2) When the equation f (x) = 0 with respect to X has a real solution, the value range of a is obtained
(3) If any x ∈ R, there is - 5 ≤ f (x) ≤ - 1, find the value range of real number a
A:
f(x)=cos²x+asinx-2a-2
=1-sin²x+asinx-2a-2
=-sin²x+asinx-2a-1
=-(sinx-a/2)²+a²/4-2a-1
1) When a = - 2:
f(x)=-(sinx+1)²+4=0
SiNx + 1 = 2 or SiNx + 1 = - 2
So: SiNx = 1 (SiNx = - 3 does not conform to rounding)
So: x = 2K π + π / 2, K ∈ Z
2）
F (x) = - Sin & # 178; X + asinx-2a-1 = 0 has real solution
Let t = SiNx ∈ [- 1,1], and the equation be reduced to: T & # 178; - at + 2A + 1 = 0
The results are as follows
a=(t²+1)/(t-2)=(t-2)+5/(t-2)+4
Given that a, B and X are positive numbers and LG (BX) · LG (AX) + 1 = 0, the value range of AB is obtained
∵ a, B, X are positive numbers, and LG (BX) · LG (AX) + 1 = 0, ∵ (LGA + lgx) (LGB + lgx) + 1 = 0, we get (lgx) 2 + (LGA + LGB) lgx + 1 + lgalgb = 0, ∵ this equation has a solution, ∵ = (LGA + LGB) 2-4lgalgb-4 ≥ 0 (LGA) 2 + 2lgalgb + (LGB) 2-4lgalgb-4 ≥ 0
If y = f (x) is a function defined on R and satisfies f (0) = 1, and for any real number x, y has the analytic expression f (X-Y) = f (x) - Y (2x-y + 1) for y = f (x)
When y = x, f (0) = f (x) - x (2x-x + 1) f (0) = 1
∴y=f(x)=x²+x+1
Let y = X
f（x-x）=f（x）-x（2x-x+1）
1=f（x）-x（x+1）
f（x）=x^2+x+1
Let A. B be a positive number, if LG (AX) LG (BX) + 1 = 0 has a solution, then the value range of a / B is obtained
(lga+lgx)(lgb+lgx)+1=0
lg²x+(lga+lgb)lgx+lgalgb+1=0 ①
If LG (AX) LG (BX) + 1 = 0, there is a solution
Then there must be lgx satisfying Formula 1
Δ≥0
(lga+lgb)²-4(lgalgb+1)≥0
(lga-lgb)²≥4
[lg(a/b)]²≥4
LG (A / b) ≥ 2 or LG (A / b) ≤ - 2
A / b ≥ 100 or 0
If f (x) is an odd function defined on R, and X ∈ (0, + ∞), f (x) = LG (x + 1), find the expression of F (x) and draw the diagram
① When x = 0, f (0) = 0; ② when x ＜ 0, - x ＞ 0, ∵ f (x) is an odd function, ∵ f (- x) = - f (x) ∵ f (x) = - f (- x) = - LG (- x + 1), to sum up: F (x) = LG (x + 1), (x ＞ 0) 0, (x = 0) − LG (− x + 1), and (x ＜ 0) the image is shown in the following figure:
Given that a, B and X are positive numbers and LG (BX) · LG (AX) + 1 = 0, the value range of AB is obtained
∵ a, B, X are positive numbers, and LG (BX) · LG (AX) + 1 = 0, ∵ (LGA + lgx) (LGB + lgx) + 1 = 0, and (lgx) 2 + (LGA + LGB) lgx + 1 + lgalgb = 0, ∵ this equation has a solution, ∵ = (LGA + LGB) 2-4lgalgb-4 ≥ 0 (LGA) 2 + 2lgalgb + (LGB) 2-4lgalgb-4 ≥ 0 (LGA LGB) 2 ≥ 4 LGA LGB ≥ 2 or LGA LGB ≤ - 2lg (a-b) ≥ 2 or LGA / b ≤ - 2 ∵ ab ≥ 100 Or 0 ＜ ab ≤ 1100. The range of AB is (01100) ∪ [100, + ∞)
Let f (x) be a function defined on the real number set R and satisfy f (x + 2) = f (x + 1) - f (x). If f (1) = Lg3 / 2, f (2) = LG15, find f (2004)
F (x + 2) = f (x + 1) - f (x) f (x + 3) = f (x + 2) - f (x + 1) = f (x + 1) - f (x) - f (x + 1) = - f (x) f (x + 6) = - f (x + 3) = f (x), so f (x) is a function with period 6, because f (2004) = f (6 * 334 + 0) = f (0) because f (2) = f (1) - f (0) LG15 = Lg3 / 2-F (0) f (0) = - 1
As soon as I see this kind of topic, I want to see if there are rules
f(1)=lg3/2
f(2)=lg15
f(3)=lg10
f(4)=lg2/3
f(5)=lg1/15
f(6)=lg1/10
f(7)=lg3/2=f(1)
f(8)=lg15=f（2）
So f (2008) = f (6 * 334 + 4) = f (4) = LG (2 / 3)
When a is a, the equation LG (AX) = 2lg (x + 1) has a solution?
The original formula can be reduced to AX = (x + 1) & sup2;, that is, X & sup2; + (2-A) x + 1 = 0

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